Two hockey pucks, each with a mass of 0.2 kg, slide across the ice and

Two charges, one of 2.50μC and the other of -3.50μC, are placed on the x-axis, one at the origin and the other at x = 0.600 m

aev [14]

Answer:

Explanation:

Given

charge of first body $q_1=2.5\ mu C$

charge of second body $q_2=-3.5\ mu C$

Particle 1 is at origin and particle 2 is at $x=0.6\ m$

third Particle which charge +q must be placed left of $2.5\mu C$ because it will repel the q charge while $-3.5\mu C$ will attract it

suppose it is placed at a distance of x m

$F_{1q}=\frac{kq(2.5)}{x^2}$

$F_{2q}=\frac{kq(-3.5)}{(0.6+x)^2}$

$F_{1q}+F_{2q}=0$

$\frac{kq(2.5)}{x^2}+\frac{kq(-3.5)}{(0.6+x)^2}=0$

$\frac{kq(2.5)}{x^2}=\frac{kq(3.5)}{(0.6+x)^2}$

$\frac{0.6+x}{x}=(\frac{3.5}{2.5})^{0.5}$

$0.6+x=1.1832x$

$x=3.27\ m$

5 0

3 years ago

An apple is whirled round in a horizontal circle on the end of a string which is tied to the stalk. It is whirled faster and fas

jek_recluse [69]

when the apple moves in a horizontal circle, the tension force in the string provides the necessary centripetal force to move in circle. the tension in the string is given as

T=mv²/r

where T = tension force in the string , m = mass of the apple

v = speed of apple , r = radius of circle.

clearly , tension force depends on the square of the speed. hence greater the speed, greater will be the tension force.

at some point , the speed becomes large enough that it makes the tension force in the string becomes greater than the tensile strength of the string. at that point , the string breaks

6 0

3 years ago

A convex spherical mirror having a radius of curvature 18 cm (focal length = 1/2 radius of curvature for a spherical mirror) pro

tekilochka [14]

Answer:

distance between object and image = 18.9 cm

Explanation:

given data

radius of curvature = 18 cm

focal length = 1/2 radius of curvature

magnification = 40%

to find out

distance between object and image

solution

we know lens formula that is

1/f = 1/v + 1/u ....................1

here f = 18 /2 and v and u is object and image distance

and we know m = 40% = 0.40

so 0.40 = -v / u

so here v = - 0.40 u

so from equation 1

1/f = 1/v + 1/u

2/18 = - 1/0.40u + 1/u

u = -13.5 cm ..................2

and

v = -0.40 (- 13.5)

v = 5.4 cm ......................3

so from equation 2 and 3

distance between object and image = 5.4 + 13.5

distance between object and image = 18.9 cm

6 0

3 years ago

A 1200-kilogram automobile in motion strikes a 0.0001-kilogram insect. As a result, the insect is accelerated at a rate of 100 m

nadya68 [22]

Answer:

force = 1 × $10^{-2}$ N

Explanation:

given data

automobile mass = 1200 kg

insect mass = 0.0001 kg

insect accelerated = 100 m/s²

to find out

magnitude of the force the insect exerts on the car

solution

we get here force the insect exerts that is express as

force = mass × acceleration ............1

put here value we get

force = 0.0001 × 100 m/s²

force = 1 × $10^{-2}$ N

6 0

3 years ago

Read 2 more answers

When going from a fast speed to a slow speed how is light bent?

Zolol [24]

Toward the normal line

4 0

3 years ago

Read 2 more answers